So the only candidates for {r2c1, r2c2, r8c1} are v=3469, the only candidates for {r1c3, r1c7, r3c3} are v=1345, and "3" is restricted common. Since r1c1 can "see" all the instances of "4" in both sets, the rule says I can exclude "4" from r1c1. Everything seems dandy.

But then I got to thinking about the proof that lies behind the "ALS XZ rule". In this instance it doesn't really apply, because I can complete the two ALS's as follows.

and no contradiction is yet apparent, because the "locked common" value, "3", can "escape" into r7c2. I have to extend the argument farther, along the lines identified as a "DIC" above, before the contradiction is forced.

I'm thinking the reason the proof fails here is that the definition

Quote:

An almost locked set consists of N cells with exactly N+1 candidates

is not quite restrictive enough. I'm not quite sure how it should be phrased, but it seems that cases like this -- where the two instances of candidate "9" in the set marked with a minus sign can't "see" each other -- need to be excluded somehow. dcb

For cells to form an ALS, they need to see each other, just like a naked or hidden subset. For this reason, r2c1, r2c2, r8c1 cannot be an ALS.

Having said that, there are some developments towards extended sets which partially overlap. Myth might be able to tell you more about them, or you could have a look at the Pappocom forum. Search for "COALS".

Ruud_________________“If the human brain were so simple that we could understand it, we would be so simple that we couldn't.” - Emerson M Pugh